18.702: Quiz 1 Solutions

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1 MIT MATHEMATICS : Quiz 1 Solutions February There are four problems on this quiz worth equal value. You may quote without proof any result stated in class or in the assigned reading, unless you are specifically asked to prove the statement in question. You may not consult outside sources while answering this quiz. You have fifty minutes. Name: Dhruv 1

2 0.1 FUNDAMENTAL NOTIONS Decide whether the following statements are true or false. No justification is necessary. All representations in this problem are finite dimensional over the complex numbers. (A) T/F If every irreducible representation of a finite group G is one-dimensional, then G is abelian. (B) T/F If ρ : G GL(n,C) is a faithful representation, then n G. (C) T/F For every positive integer n, there exists a group G with exactly n distinct (i.e nonisomorphic) irreducible representations. (D) T/F (E) T/F Every representation of S n is determined by its value on transpositions. Every character of S n is determined by its value on transpositions. (A) True The number of conjugacy classes is equal to the number of irreducible representations. If the sum of the squares of the dimensions of these representations has to be equal to the size of G, and each one is one-dimensional, there must be as much group elements as there are conjugacy classes, so G is abelian. (B) False Take G to be any cyclic group of finite order n and take G GL(1,C) sending the generator to a a primitive nth root of unity. (C) True Similarly to the first question, we equivalently want to find a group with n conjugacy classes and the cyclic group of order n has this number. (D) True A representation ρ : S n GL(W ) is a homomorphism, and thus is determined by its value on any generating set, in particular, the transpositions. (E) False! A character is not a representation in general. If this were true, it would mean that the space of class functions is 1-dimensional, which it is clearly not once n 3. 2

3 0.2 REAL REPRESENTATIONS Construct an irreducible two dimensional representation of the cyclic group C 3 over the real numbers, i.e. a homomorphism C 3 GL(2,R) with no stable subspaces. Say C 3 = 1, g, g 2. A representation ρ is fully determined by the choice of g, and any choice can be made provided ρ(g 3 ) = I d. Thus, we may choose ρ(g ) to be the rotation matrix with angle 2π/3. It is irreducible because the action divides up R 2 into three sectors, and none of the sectors contains a line. Said differently, there does not exist any nonzero vector that is sent to a linear multiple of itself by g. Let C n be the cyclic group on n elements. Describe all one-dimensional real representations of C n. A one-dimensional irreducible representation is a homomorphism ρ : C n GL(1,R) = R. Since ρ is a homomorphism, it must send a generator g to an nth root of unity. If n is odd, the only real root is 1, and if n is even there are two real roots, 1 and 1. 3

4 0.3 CONJUGATE REPRESENTATIONS Let G be a finite subgroup of GL(n,C). Prove that G is conjugate to a subgroup of unitary matrices. The inclusion homomorphism G GL(n,C) is a representation. It is equivalent to a unitary representation as in the proof of Maschke s theorem, choose a G-invariant Hermitian inner product, G for G by averaging any inner product over the group. For this inner product, G is equivalent to a unitary representation, i.e. it is unitary in a choice of basis for C n that is orthonormal for the inner product, G. The change of basis matrix from the original basis to this new orthonormal one conjugates G to a subgroup of the unitary group. Is it true that every finite subgroup of GL(n,R) is conjugate to a subgroup of orthogonal matrices? Give a yes/no answer and a short (2-3 sentence) justification. It most certainly is true. Repeat the proof above replacing the Hermitian inner product above by a symmetric bilinear inner product that is G-invariant. This is done identically, by starting with any old dot product and averaging over the group. The change of basis matrix conjugates G into a subgroup of orthogonal matrices. 4

5 0.4 KERNELS OF CHARACTERS Let χ : G C be a character of a finite dimensional representation of a finite group G. The kernel of χ is defined as ker(χ) = {g G : χ(g ) = χ(e)}, where e is the identity element of G. Prove that if g lies in the kernel of every irreducible character χ of G, then g must be the identity. Let g G. If g is different from the identity, then in particular, they are in different conjugacy classes, since the identity commutes with every element in G and thus forms its own conjugacy class. Thus, there is a class function that tells apart g and e. Specifically, choose the class function f : G C that is zero on e and 1 on the class of g. However, the irreducible characters form an orthonormal basis for the class functions, so there must exist an irreducible character taking different values on e and g (Why is this true?). In short, if a class function can tell g and e apart, then a character can too, because the characters form a basis for all class functions. Since this is not the case, e and g must in fact be the same element! 5

4 Group representations

Physics 9b Lecture 6 Caltech, /4/9 4 Group representations 4. Examples Example : D represented as real matrices. ( ( D(e =, D(c = ( ( D(b =, D(b =, D(c = Example : Circle group as rotation of D real vector